1. Calculate the trend surface for higher order polynomials. Do higher order polynomials capture large scale trends or are all the trends finer scale? What order polynomial would you choose to detrend your data?
From these plots, I’d choose a 4 degree polynomial - it appears to capture enough fine scale trends.
2. EXTRA CREDIT: There are two other correlation functions built into the spatial package, the Gaussian covariance gaucov and the spherical covariance sphercov. Derive MLE for these two functions via numerical optimization. Turn in an AIC table for the models you fit and identify the best fitting model that you will use for Kriging
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3. Explain the figure generated from the previous box of code. What is being plotted (axes, data, model)? How would you interpret this figure? How would you interpret the parameters of the best-fit model? What is the range? Note: you do not need to include the Gaussian and spherical lines if you did not fit them in the previous step.
This is a plot of the autocorrelation of the data.
In this case, I like the Gaussian fit the best because
4. Include your kriged map of mean US ozone for 2008 in your lab report. Where in the country has the highest ozone levels?
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Generate a second map of mean US ozone but this time include contours based on the error matrix rather than the prediction matrix. Where are ozone levels most uncertain?
.5 There are a few reasons that one might be skeptical of this initial exploratory map. Name at least two and describe ways the model could be improved (without collecting additional data).
We did not explore the possibility of anistropy. It is possible that the spacial covariance is not the same in all directions.
The variogram does not fit the standard curve we had looked at before - it doesn’t converge nicely to the sill and has a large nugget value.
My initial choice for the order of the polynomial of the trend was never proven to be the “correct” ie optimal value.
7. Include the time series plot and briefly describe the features of the overall trend in ozone concentrations.
The data is trending downwards in what could be fitted with a line, however, the moving average and lowess curve seem to show an ocilation that might be better fit with a higher order polynomial.
8. Does the detrended data meet the assumptions of second order stationarity? Why or why not?
Overall, I would be warry of saying this is stationary, while the mean is close to 0, the histogram does appeared to be skewed to the right - this can also be seen on both sides, expecially the right hand side of the Q-Q plot. Thus the homeskedastic variance will be non-zero. It would be worth doing a reanalysis in which we identify potential outliers.
9. Does the first-difference time series meet the assumptions of second order stationarity? Why or why not?
Once again, I’m skeptical because even thought the mean residual error near 0, there is a bias present that that results in a non-zero homeskedastic variance. Also, even without the bias, the histogram doens’t make a very clean normal curve.
Autocorrelation and partial autocorrelation in the time series:
12. Based on the diagnostics performed, what ARIMA model is likely to perform best? What orders should p, d, and q be? Why? Should you fit the model to xt or rx?
13. Fit the arima model you proposed using the function arima:
# plot(rx)
# abline(h=0,lty=2)
# hist(rx,10)
#
# plot(diff(rx))
# abline(h=0,lty=2)
# hist(diff(rx,10))
#
# acf(diff(rx))
# pacf(diff(rx))
#
p <- 0
d <- 1
q <- 0
arima(rx,c(0,1,1)) # my guess - aic = -225.47
Then propose alternative models that are similar to the one you fit (e.g. increase or decrease orders by 1). Based on AIC scores what model provides the best fit? Provide a table of models you tried and their AIC scores.
## <!-- html table generated in R 3.1.1 by xtable 1.7-4 package -->
## <!-- Sun Dec 14 15:41:10 2014 -->
## <table border=1>
## <tr> <th> </th> <th> V1 </th> <th> V2 </th> </tr>
## <tr> <td align="right"> 1 </td> <td> p= 0,d=1,q=1 | </td> <td> -225.65578896828 </td> </tr>
## <tr> <td align="right"> 2 </td> <td> p= 0,d=0,q=0 | </td> <td> -238.362985530861 </td> </tr>
## <tr> <td align="right"> 3 </td> <td> p= 0,d=1,q=0 | </td> <td> -209.766255142534 </td> </tr>
## <tr> <td align="right"> 4 </td> <td> p= 0,d=1,q=2 | </td> <td> -223.811783223323 </td> </tr>
## <tr> <td align="right"> 5 </td> <td> p= 1,d=1,q=1 | </td> <td> -223.700548251881 </td> </tr>
## <tr> <td align="right"> 6 </td> <td> p= 2,d=1,q=1 | </td> <td> -225.876448072165 </td> </tr>
## <tr> <td align="right"> 7 </td> <td> p= 0,d=0,q=1 | </td> <td> -236.944190351403 </td> </tr>
## </table>